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Unformatted text preview: 0.1 0.2 Solution to Exercises 1. Which of the following are Brownian motions? (a) { √ tB 1 ,t ≥ } ; (b) { B 2 t B t ,t ≥ } . Solution. (a) Write X t = √ tB 1 . Clearly, for s &lt; t E ( X t ) = 0 Cov ( X s ,X t ) = Cov ( √ sB 1 , √ tB 1 ) = √ stEB 2 1 = √ st 6 = s. Thus, { √ tB 1 ,t ≥ } is NOT a Brownian motion. (b) Write Y t = B 2 t B t . Clearly, for s &lt; t , we have E ( Y t ) = 0 Cov ( Y s ,Y t ) = Cov ( B 2 s B s ,B 2 t B t ) = E ( B 2 s B 2 t ) + E ( B s B t ) E ( B 2 s B t ) E ( B s B 2 t ) = 2 s + s min(2 s,t ) s = 2 s min(2 s,t ) , which is in general not equal to s . For instance, if 0 &lt; 2 s &lt; t , then Cov ( Y s ,Y t ) = 2 s 2 s = 0 6 = s . Thus, { B 2 t B t ,t ≥ } is NOT a Brownian motion. 2. What is the distribution of B s + B t , s ≤ t ? Solution. Solution. E ( B s + B t ) = 0, var ( B s + B t ) = var ( B s )+ var ( B t )+2 E ( B s B t ) = s + t +2 s = 3 s + t . So B s + B t ∼ N (0 , 3 s + t ). 3. Compute E [ B t 1 B t 2 B t 3 ] for t 1 &lt; t 2 &lt; t 3 . Solution . Let with 0 = t ≤ t 1 &lt; ... &lt; t n = 1 and B t i ’s are independent N (0 , Δ t i ), and E [ B t 1 ( B t 2 + B t 2 )( B t 1 + B t 2 + B t 3 )] = EB 3 t 1 = 0. 4. Calculate P ( B ( t ) ≤ 0 for t = 2) and P ( B (1) ≤ ,B (2) ≤ 0). Solution. P ( B (2) ≤ 0) = 1 / 2 as B (2) ∼ N (0 , 2). Next, Let X = B (1) and Y = B (2) B (1). Then X and Y are i.i.d. N (0 , 1). Hence, P ( B (1) ≤ ,B (2) ≤ 0) = P ( X ≤ ,X + Y ≤ 0) = Z ∞ x =∞ P ( X ≤ 0; X + Y ≤  X = x ) d Φ( x ) = Z x =∞ P ( Y ≤  x  X = x ) d Φ( x ) = Z x =∞ Φ( x ) d Φ( x ) = Z x =∞ [1 Φ( x )] d Φ( x ) = Z 1 / 2 t =0 [1 t ] dt = 3 / 8 . 5. Let Y 1 ,Y 2 be independent N (0 , 1) r.v.’s and for some constant w set X t = Y 1 cos( wt ) + Y 2 sin( wt ) ,∞ &lt; t &lt; ∞ . Show that (a) { X t } is a weakly stationary process; (b) { X t } is a (strictly) stationary process. 1 Solution. • EX t = EY 1 cos ( wt ) + EY 2 sin ( wt ) = 0, and let s &lt; t , then Cov ( X s ,X t ) = E ( X s X t ) = E [ Y 1 cos ( ws ) + Y 2 sin ( ws )][ Y 1 cos ( wt ) + Y 2 sin ( wt )] = cos ( ws ) cos ( wt ) + sin ( ws ) sin ( wt ) = cos (( t s ) w ) • Since X t is a Gaussian process, weak and strict stationarities are equivalent. 6. Suppose that S ( t ), t ≥ 0, is a geometric Brownian motion with drift parameter μ = . 01 and volatility parameter σ = 0 . 2. If S (0) = 100, find • ES (10), • P ( S (10) &gt; 100), • P ( S (10) &lt; 110). Solution. S t = S e μt + σB t = 100 e . 01 t +0 . 2 B t , and ES t = S e μt +0 . 5 σ 2 t = 100 e . 03 t . • ES 10 = 100 e . 3 = 134 . 99. • P ( S 10 &gt; 100) = P (100 e . 1+0 . 2 B 10 &gt; 100) = P ( B 10 &gt; 1 / 2) = P ( B 1 &gt; . 5 / √ 10) = . 564....
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This note was uploaded on 02/01/2012 for the course MATH 5010 taught by Professor D during the Spring '11 term at HKU.
 Spring '11
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